In 3D rendering (or geometry for that matter), in the rasterization algorithm, when you project the vertices of a triangle onto the screen and then find if a pixel overlaps the 2D triangle, you often need to find the depth or the z-coordinate of the triangle that the pixel overlaps. Generally, the method consists of computing the barycentric coordinates of the pixel in the 2D "projected" image of the triangle, and then use these coordinates to interpolate the triangle original vertices z-coordinates (before the vertices got projected).
Now it's written in all text books that you can't interpolate the vertices coordinates of the vertices directly but that you need to do this instead:
(sorry can't get Latex to work?)
1/z = w0 * 1/v0.z + w1 * 1/v1.z + w2 * 1/v2.z
Where w0, w1, and w2 are the barycentric coordinates of the "pixel" on the triangle.
Now, what I am looking after, are two things:
what would be the formal proof to show that interpolating z doesn't work?
what would be the formal proof to show that 1/z does the right thing?
To show this is not home work ;-) and that I have made some work on my own, I have found the following explanation for question 2.
Basically a triangle can be defined by a plane equation. Thus you can write:
Ax + By + Cz = D.
Then you isolate z to get z = (D - Ax - By)/C
Then you divide this formula by z as you would with a perspective divide and if you develop, regroup, etc. you get:
1/z = C/D + A/Dx/z + B/Dy/z.
Then we name C'=C/D B'=B/D and A'=A/D you get:
1/z = A'x/z + B'y/z + C'
It says that x/z and y/z are just the coordinates of the points on the triangles once projected on the screen and that the equation on the right is an "affine" function therefore 1/z is a linear function???
That doesn't seem like a demonstration to me? Or maybe it's the right idea, but can't really say how you can tell by just looking at the equation that this is an affine function. If you multiply all the terms you just get:
A'x + B'y + C'z = 1.
Which is just basically our original equations (just need to replace A' B' and C' with the proper term).
Not sure what you are trying to ask here, but if you look at:
1/z = A'x/z + B'y/z + C'
and rewrite it as:
1/z = A'u + B'v + C'
where (u,v) are screen coordinates of the triangle after perspective projection, you can see that the depth (z) of a point on the triangle is not linearly related to (u,v) but 1/depth is and that is what the textbooks are trying to teach you.
Related
Ive got 2 points in 3d space (with the same y coordinate). Ill call them c and m. I want to find the corner points (marked in the pic as p1-p4) of a square with the width w. The important thing is, that the square is not parallel to the x-axis. If it were, (for p1 as an example) I could just do:
p1.x = m.x + w / 2
p1.y = m.y + w / 2
p1.z = m.z
How would I do the same with a angled square? These are all the given points:
m; c
and lenghts:
w; d
There's multiple ways to do it, but here's one way.
If the two points are guaranteed to have the same y value, you should be able to do it as follows.
Take 'm - c' and call that u. Normalize u. Then take the cross product of u and the y axis to get v, a vector parallel to the xz plane that's perpendicular to u. (This can be optimized, but that's unlikely to be important.) Then take the cross product of u and v to get a third vector, w. Note that you can use 'm - c' or 'c - m', or use different orders for the cross-product arguments, and it'll still work, but the resulting vectors may point in different directions (but only opposite directions). You can also normalize at different points in the process and get the same results at the end.
Once you have m, v, and w, you can use some basic vector math to compute the corners.
[Edit: I see you have a variable named 'w', so I should clarify that the 'w' in my example is a different 'w' than yours. As for your 'w' and 'd', those would factor in in the vector math I mentioned at the end.]
I think there is a pretty straightForward answer to this, but i cant find it. My geometry lessons are too far away for this. the problem is:
Given 2 points A and B (coordinates Ax Ay Bx and By), I want to find the coordinates of points C and D so that [AB] and [CD] segments intersect at their center and [CD] has a length of d (a variable).
I want to find the equation giving me Cx,Cy,Dx and Dy from Ax,Ay,Bx,By and d.
Here is a little schema of the problem:
and an image of the intended result:
I already know how to find the center point of [AB] (Ax+Bx/2, Ay+By/2), how to find the slope of the [AB] segment (By-Ay/Bx-Ax) and then one of the [CD] segment (Ax-Bx/By-Ay). But then i get stuck on how to get my two points. I thought i could calculate the angle from the slope, then use it with some trigonometry to get the coordinates but it sounds like a quite heavy, ugly and unnecessary calculation...
It feels so close, but i still cant get it.
I also found this post, which is almost perfect, but the length cannot be defined: it must be the same as the first segment.
I dont think this is language-dependent, but if you must know, i'm doing a mini prototype on processing and will probably get it on javascript later.
Thanks for any help.
The basic trick here is that, in 2d, the perpendicular to a vector (x, y) is merely ± (-y, x). (One gets this by computing the cross product with the (0, 0, 1) vector in 3d, and projecting to 2d.) So what you need to do is:
Get the midpoint between A and B (you have done that).
Get the vector from A to B, which is B - A = (x, y) = (bx - ax, by - ay).
Get the perpendicular vector: (-y, x).
Normalize it. Let length = sqrt(y*y + x*x), then norm = (-y/length, x/length).
Multiply the normalized perpendicular by your desired distance ± d/2 (since you want the distance between C and D to be d), and add to the center point.
No slopes or trig functions are required.
I am doing ray tracing with sphere. When we solve quadratic equation and get two roots then which value we need to take? My understanding is that we need to take smallest value. But with that i get wrong results when i put that value in calculating intersection point and further lighting calculations. When I take larger one it gives correct results. I saw some examples on the net and some of them are taking smaller one and others are taking larger. I am really confused which one is correct and why.
Suppose the origin of the ray is O and the direction is R (ideally a unit vector). Then the ray is defined by the parametric equation O + tR. Break this into components Ox + tRx, Oy + tRy and Oz + tRz. Substitute these into the equation for the circle x^2 + y^2 + z^2 = 0. This produces a quadratic equation with the variable t. Find the roots of the equation.
Now, if you have no real roots, the ray doesn't hit the sphere. If you have one real root then you hit the side of the sphere. If you have two real roots, then you hit the sphere twice. You want the closer root (the smaller t) unless t is negative in which case the intersection point is behind you and you don't care. If both roots are negative, then both intersection points are behind the origin of the ray and you can ignore them.
If the sphere is transformed away from the origin and you care about rotation and scaling of the sphere, then the equation of the sphere becomes more complex with xy, yz, xz, x, y and z terms. A general quadric shape is defined as Ax^2 + By^2 + Cy^2 + Dxy + Eyz + Fxz + Gx + Hy + Jz + K = 0. The principle above still applies. Substitute the parametric line equation in, solve for t, take the smallest non-negative root.
Your intersection point is O+tR where t is the smallest non-negative root. From there, I don't know why your lighting calculations would be off.
I have the plane equation describing the points belonging to a plane in 3D and the origin of the normal X, Y, Z. This should be enough to be able to generate something like a 3D arrow. In pcl this is possible via the viewer but I would like to actually store those 3D points inside the cloud. How to generate them then ? A cylinder with a cone on top ?
To generate a line perpendicular to the plane:
You have the plane equation. This gives you the direction of the normal to the plane. If you used PCL to get the plane, this is in ModelCoefficients. See the details here: SampleConsensusModelPerpendicularPlane
The first step is to make a line perpendicular to the normal at the point you mention (X,Y,Z). Let (NORMAL_X,NORMAL_Y,NORMAL_Z) be the normal you got from your plane equation. Something like.
pcl::PointXYZ pnt_on_line;
for(double distfromstart=0.0;distfromstart<LINE_LENGTH;distfromstart+=DISTANCE_INCREMENT){
pnt_on_line.x = X + distfromstart*NORMAL_X;
pnt_on_line.y = Y + distfromstart*NORMAL_Y;
pnt_on_line.z = Z + distfromstart*NORMAL_Z;
my_cloud.points.push_back(pnt_on_line);
}
Now you want to put a hat on your arrow and now pnt_on_line contains the end of the line exactly where you want to put it. To make the cone you could loop over angle and distance along the arrow, calculate a local x and y and z from that and convert them to points in point cloud space: the z part would be converted into your point cloud's frame of reference by multiplying with the normal vector as with above, the x and y would be multiplied into vectors perpendicular to this normal vectorE. To get these, choose an arbitrary unit vector perpendicular to the normal vector (for your x axis) and cross product it with the normal vector to find the y axis.
The second part of this explanation is fairly terse but the first part may be the more important.
Update
So possibly the best way to describe how to do the cone is to start with a cylinder, which is an extension of the line described above. In the case of the line, there is (part of) a one dimensional manifold embedded in 3D space. That is we have one variable that we loop over adding points. The cylinder is a two dimensional object so we have to loop over two dimensions: the angle and the distance. In the case of the line we already have the distance. So the above loop would now look like:
for(double distfromstart=0.0;distfromstart<LINE_LENGTH;distfromstart+=DISTANCE_INCREMENT){
for(double angle=0.0;angle<2*M_PI;angle+=M_PI/8){
//calculate coordinates of point and add to cloud
}
}
Now in order to calculate the coordinates of the new point, well we already have the point on the line, now we just need to add it to a vector to move it away from the line in the appropriate direction of the angle. Let's say the radius of our cylinder will be 0.1, and let's say an orthonormal basis that we have already calculated perpendicular to the normal of the plane (which we will see how to calculate later) is perpendicular_1 and perpendicular_2 (that is, two vectors perpendicular to each other, of length 1, also perpendicular to the vector (NORMAL_X,NORMAL_Y,NORMAL_Z)):
//calculate coordinates of point and add to cloud
pnt_on_cylinder.x = pnt_on_line.x + 0.1 * perpendicular_1.x * 0.1 * cos(angle) + perpendicular_2.x * sin(angle)
pnt_on_cylinder.y = pnt_on_line.y + perpendicular_1.y * 0.1 * cos(angle) + perpendicular_2.y * 0.1 * sin(angle)
pnt_on_cylinder.z = pnt_on_line.z + perpendicular_1.z * 0.1 * cos(angle) + perpendicular_2.z * 0.1 * sin(angle)
my_cloud.points.push_back(pnt_on_cylinder);
Actually, this is a vector summation and if we were to write the operation as vectors it would look like:
pnt_on_line+perpendicular_1*cos(angle)+perpendicular_2*sin(angle)
Now I said I would talk about how to calculate perpendicular_1 and perpendicular_2. Let K be any unit vector that is not parallel to (NORMAL_X,NORMAL_Y,NORMAL_Z) (this can be found by trying e.g. (1,0,0) then (0,1,0)).
Then
perpendicular_1 = K X (NORMAL_X,NORMAL_Y,NORMAL_Z)
perpendicular_2 = perpendicular_1 X (NORMAL_X,NORMAL_Y,NORMAL_Z)
Here X is the vector cross product and the above are vector equations. Note also that the original calculation of pnt_on_line involved a vector dot product and a vector summation (I am just writing this for completeness of the exposition).
If you can manage this then the cone is easy just by changing a couple of things in the double loop: the radius just changes along its length until it is zero at the end of the loop and in the loop distfromstart will not start at 0.
Given two image buffers (assume it's an array of ints of size width * height, with each element a color value), how can I map an area defined by a quadrilateral from one image buffer into the other (always square) image buffer? I'm led to understand this is called "projective transformation".
I'm also looking for a general (not language- or library-specific) way of doing this, such that it could be reasonably applied in any language without relying on "magic function X that does all the work for me".
An example: I've written a short program in Java using the Processing library (processing.org) that captures video from a camera. During an initial "calibrating" step, the captured video is output directly into a window. The user then clicks on four points to define an area of the video that will be transformed, then mapped into the square window during subsequent operation of the program. If the user were to click on the four points defining the corners of a door visible at an angle in the camera's output, then this transformation would cause the subsequent video to map the transformed image of the door to the entire area of the window, albeit somewhat distorted.
Using linear algebra is much easier than all that geometry! Plus you won't need to use sine, cosine, etc, so you can store each number as a rational fraction and get the exact numerical result if you need it.
What you want is a mapping from your old (x,y) co-ordinates to your new (x',y') co-ordinates. You can do it with matrices. You need to find the 2-by-4 projection matrix P such that P times the old coordinates equals the new co-ordinates. We'll assume that you're mapping lines to lines (not, for instance, straight lines to parabolas). Because you have a projection (parallel lines don't stay parallel) and translation (sliding), you need a factor of (xy) and (1), too. Drawn as matrices:
[x ]
[a b c d]*[y ] = [x']
[e f g h] [x*y] [y']
[1 ]
You need to know a through h so solve these equations:
a*x_0 + b*y_0 + c*x_0*y_0 + d = i_0
a*x_1 + b*y_1 + c*x_1*y_1 + d = i_1
a*x_2 + b*y_2 + c*x_2*y_2 + d = i_2
a*x_3 + b*y_3 + c*x_3*y_3 + d = i_3
e*x_0 + f*y_0 + g*x_0*y_0 + h = j_0
e*x_1 + f*y_1 + g*x_1*y_1 + h = j_1
e*x_2 + f*y_2 + g*x_2*y_2 + h = j_2
e*x_3 + f*y_3 + g*x_3*y_3 + h = j_3
Again, you can use linear algebra:
[x_0 y_0 x_0*y_0 1] [a e] [i_0 j_0]
[x_1 y_1 x_1*y_1 1] * [b f] = [i_1 j_1]
[x_2 y_2 x_2*y_2 1] [c g] [i_2 j_2]
[x_3 y_3 x_3*y_3 1] [d h] [i_3 j_3]
Plug in your corners for x_n,y_n,i_n,j_n. (Corners work best because they are far apart to decrease the error if you're picking the points from, say, user-clicks.) Take the inverse of the 4x4 matrix and multiply it by the right side of the equation. The transpose of that matrix is P. You should be able to find functions to compute a matrix inverse and multiply online.
Where you'll probably have bugs:
When computing, remember to check for division by zero. That's a sign that your matrix is not invertible. That might happen if you try to map one (x,y) co-ordinate to two different points.
If you write your own matrix math, remember that matrices are usually specified row,column (vertical,horizontal) and screen graphics are x,y (horizontal,vertical). You're bound to get something wrong the first time.
EDIT
The assumption below of the invariance of angle ratios is incorrect. Projective transformations instead preserve cross-ratios and incidence. A solution then is:
Find the point C' at the intersection of the lines defined by the segments AD and CP.
Find the point B' at the intersection of the lines defined by the segments AD and BP.
Determine the cross-ratio of B'DAC', i.e. r = (BA' * DC') / (DA * B'C').
Construct the projected line F'HEG'. The cross-ratio of these points is equal to r, i.e. r = (F'E * HG') / (HE * F'G').
F'F and G'G will intersect at the projected point Q so equating the cross-ratios and knowing the length of the side of the square you can determine the position of Q with some arithmetic gymnastics.
Hmmmm....I'll take a stab at this one. This solution relies on the assumption that ratios of angles are preserved in the transformation. See the image for guidance (sorry for the poor image quality...it's REALLY late). The algorithm only provides the mapping of a point in the quadrilateral to a point in the square. You would still need to implement dealing with multiple quad points being mapped to the same square point.
Let ABCD be a quadrilateral where A is the top-left vertex, B is the top-right vertex, C is the bottom-right vertex and D is the bottom-left vertex. The pair (xA, yA) represent the x and y coordinates of the vertex A. We are mapping points in this quadrilateral to the square EFGH whose side has length equal to m.
Compute the lengths AD, CD, AC, BD and BC:
AD = sqrt((xA-xD)^2 + (yA-yD)^2)
CD = sqrt((xC-xD)^2 + (yC-yD)^2)
AC = sqrt((xA-xC)^2 + (yA-yC)^2)
BD = sqrt((xB-xD)^2 + (yB-yD)^2)
BC = sqrt((xB-xC)^2 + (yB-yC)^2)
Let thetaD be the angle at the vertex D and thetaC be the angle at the vertex C. Compute these angles using the cosine law:
thetaD = arccos((AD^2 + CD^2 - AC^2) / (2*AD*CD))
thetaC = arccos((BC^2 + CD^2 - BD^2) / (2*BC*CD))
We map each point P in the quadrilateral to a point Q in the square. For each point P in the quadrilateral, do the following:
Find the distance DP:
DP = sqrt((xP-xD)^2 + (yP-yD)^2)
Find the distance CP:
CP = sqrt((xP-xC)^2 + (yP-yC)^2)
Find the angle thetaP1 between CD and DP:
thetaP1 = arccos((DP^2 + CD^2 - CP^2) / (2*DP*CD))
Find the angle thetaP2 between CD and CP:
thetaP2 = arccos((CP^2 + CD^2 - DP^2) / (2*CP*CD))
The ratio of thetaP1 to thetaD should be the ratio of thetaQ1 to 90. Therefore, calculate thetaQ1:
thetaQ1 = thetaP1 * 90 / thetaD
Similarly, calculate thetaQ2:
thetaQ2 = thetaP2 * 90 / thetaC
Find the distance HQ:
HQ = m * sin(thetaQ2) / sin(180-thetaQ1-thetaQ2)
Finally, the x and y position of Q relative to the bottom-left corner of EFGH is:
x = HQ * cos(thetaQ1)
y = HQ * sin(thetaQ1)
You would have to keep track of how many colour values get mapped to each point in the square so that you can calculate an average colour for each of those points.
I think what you're after is a planar homography, have a look at these lecture notes:
http://www.cs.utoronto.ca/~strider/vis-notes/tutHomography04.pdf
If you scroll down to the end you'll see an example of just what you're describing. I expect there's a function in the Intel OpenCV library which will do just this.
There is a C++ project on CodeProject that includes source for projective transformations of bitmaps. The maths are on Wikipedia here. Note that so far as i know, a projective transformation will not map any arbitrary quadrilateral onto another, but will do so for triangles, you may also want to look up skewing transforms.
If this transformation has to look good (as opposed to the way a bitmap looks if you resize it in Paint), you can't just create a formula that maps destination pixels to source pixels. Values in the destination buffer have to be based on a complex averaging of nearby source pixels or else the results will be highly pixelated.
So unless you want to get into some complex coding, use someone else's magic function, as smacl and Ian have suggested.
Here's how would do it in principle:
map the origin of A to the origin of B via a traslation vector t.
take unit vectors of A (1,0) and (0,1) and calculate how they would be mapped onto the unit vectors of B.
this gives you a transformation matrix M so that every vector a in A maps to M a + t
invert the matrix and negate the traslation vector so for every vector b in B you have the inverse mapping b -> M-1 (b - t)
once you have this transformation, for each point in the target area in B, find the corresponding in A and copy.
The advantage of this mapping is that you only calculate the points you need, i.e. you loop on the target points, not the source points. It was a widely used technique in the "demo coding" scene a few years back.